Applications of Differential equations

[Squires]

The rate of capital growth in a bank account is described by the differential equation

dM/dt = aM

Where dM/dt is the rate of change of the capital M and a is the annual interest rate.

Show that the general solution for the time dependence of capital M(t) is given by:

M(t) = M0 * e^at

where t is the time in years and M0 is the initial capital.

Really struggle with questions written like this, I know I should be approaching the question by integrating dM/dt = aM, but I don't understand where M0 comes into this atall, any help much appreciated!

 

[Kreizhn]

This is a differential equation, and so we want to apply differential techniques to it. In particular, might I suggest separation of variables?

Namely, if you have an equation
\frac{dM}{dt} = aM
try "multiplying" everything by
\frac{dt}{M}.

You'll then be in a position to integrate to find a solution. Don't forget your constant (hint for M_0.)

 

[Squires]

Thankyou for responding.

Yeah I understand how integrating will give me constants that show how it proves the equation, but am really unfamilliar with the differential techniques and how to get that far?

Would it be complicated to explain how to separate the variables in this instance? No worries if it's too much hassle.

 

[Kreizhn]

No problem. While mathematically it pains me to say this, you should treat \frac{dM}{dt} as a fraction so that you can separate the dM and the dt components.

Now what do these terms look like? They look like integration terms!

\int f(x) \underbrace{dx}_{\uparrow}

Here the dx terms tell us that we're integrating with respect to x. Hence our goal with

\frac{dM}{dt} = a M

will be to move all the "M" terms to one-side of the equal sign, and all the "t" terms to the other. Multiply both sides by dt and divide both sides by M, as if you were cancelling denominators in fractions

<br /> \begin{align*}\left( \frac{\cancel{dt}}{M} \right) \frac{dM}{\cancel{dt}} &amp;= \left( \frac{dt}{\cancel M} \right)(a \cancel M) \\<br /> \frac{dM}M &amp;= a dt \end{align*}<br />

Now these look like integrations right? So trying throwing an integral sign out front

\int \frac1M dM = a \int dt

Try integrating these equations (not forgetting your constant of integration!) and solve for M.

 

[Squires]

Ahaaa, maybe not quite so much detail was needed, sorry for your pain!

Thanks loads, you've made it make far more sense than any website I've looked at, and helped my coursework mark for sure, Thanks again man :)